hw5soln_f09

# hw5soln_f09 - ME 132 Fall 2009 Solutions to Homework 5 1. (...

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Unformatted text preview: ME 132 Fall 2009 Solutions to Homework 5 1. ( § 6.6, Problem 2) (a) Including the measurement noise, the closed-loop system equation becomes ˙ v ( t ) =- α m v ( t ) + E m u ( t ) + G m w ( t ) =- α m v ( t ) + E m [ K ff v des ( t )- K fb ( v ( t ) + n ( t ))] + G m w ( t ) =- α + EK fb m v ( t ) + EK ff m v des ( t ) + G m w ( t )- EK fb m n ( t ) . The two outputs are y 1 ( t ) = v ( t ) , y 2 ( t ) = u ( t ) =- K fb v ( t ) + K ff v des ( t )- K fb n ( t ) . The coefficients are then A =- α + EK fb m , B 1 = EK ff m , B 2 = G m , B 3 =- EK fb m , C 1 = 1 , D 11 = 0 , D 12 = 0 , D 13 = 0 , C 2 =- K fb , D 21 = K ff , D 22 = 0 , D 23 =- K fb . (b) The closed-loop system is stable if A < ⇔ - α + EK fb m < ⇔ α + EK fb > ⇔ K fb >- α E . (c) To determine the steady-state gain from v des to v , we set v des to a constant value v des ( t ) ≡ ¯ v des , and set the other inputs to zero, w ( t ) ≡ 0, n ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 1 ¯ v des , which has the solution v ( t ) = e At v (0) + integraldisplay t e A ( t − τ ) B 1 ¯ v des d τ = e At v (0)- B 1 A ( 1- e At ) ¯ v des . Assuming the closed-loop system is stable, the steady-state value of v ( t ) is ¯ v =- B 1 A ¯ v des . 1 Therefore, the steady-state gain from v des to v is G v des → v (0) =- B 1 A = EK ff α + EK fb . There is also a “shortcut” method of determining the steady-state gain. If the closed-loop system is stable, then we know that v ( t ) does indeed reach some steady- state value ¯ v , given constant inputs. Then we can take the limit of the closed-loop system equation as t → ∞ , 0 = A ¯ v + B 1 ¯ v des , since lim t →∞ ˙ v ( t ) = ˙ ¯ v = 0. We then obtain the same result as above, namely, ¯ v =- B 1 A ¯ v des . It is important to ensure that the closed-loop system is stable before using this shortcut method. In another problem ( § 4.5, Problem 4), we can see an example where using the shortcut method on an unstable system can lead to incorrect conclusions. (d) To determine the steady-state gain from w to v , we set w to a constant value w ( t ) ≡ ¯ w , and set the other inputs to zero, v des ( t ) ≡ 0, n ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 2 ¯ w. The steady-state value of v ( t ) is ¯ v =- B 2 A ¯ w. Therefore, the steady-state gain from w to v is G w → v (0) =- B 2 A = G α + EK fb . (e) Finally, to determine the steady-state gain from n to v , we set n to a constant value n ( t ) ≡ ¯ n , and set the other inputs to zero, v des ( t ) ≡ 0, w ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 3 ¯ n....
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## This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at Berkeley.

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hw5soln_f09 - ME 132 Fall 2009 Solutions to Homework 5 1. (...

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